Pre-Algebra

10-5 Solving for a Variable10-5 Solving for a Variable

Pre-Algebra

Warm UpWarm Up

Problem of the DayProblem of the Day

Lesson PresentationLesson Presentation

Pre-Algebra

10-5 Solving for a Variable

Warm UpSolve.

1. 8x – 9 = 23

2. 9x + 12 = 4x + 37

3. 6x – 8 = 7x + 3

4. x + 3 =

x = 4

x = 5

x = –11

Pre-Algebra

10-5 Solving for a Variable

12

18

23 4x = – , or –5 3

4

Pre-Algebra

10-5 Solving for a Variable

Problem of the Day

The formula A = 4r2 gives the surface area of a geometric figure. Solve the formula for r.

Can you identify what the geometric figure is?sphere

Pre-Algebra

10-5 Solving for a Variable

Learn to solve an equation for a variable.

Pre-Algebra

10-5 Solving for a Variable

If an equation contains more than one variable, you can sometimes isolate one of the variables by using inverse operations. You can add and subtract any variable quantity on both sides of an equation.

Pre-Algebra

10-5 Solving for a Variable

Solve for the indicated variable.

Additional Example 1A: Solving for a Variable by Addition or Subtraction

A. Solve a – b + 1 = c for a.

a – b + 1 = c

+ b – 1 + b – 1 Add b and subtract 1 from both sides.

a = c + b – 1 Isolate a.

Pre-Algebra

10-5 Solving for a Variable

Solve for the indicated variable.

Additional Example 1B: Solving for a Variable by Addition or Subtraction

B. Solve a – b + 1 = c for b.

a – b + 1 = c– a – 1 – a – 1 Subtract a and 1 from both

sides.

–b = c – a – 1 Isolate b.

Multiply both sides by –1.–1 (–b) = –1 (c – a – 1)

b = –c + a + 1 Isolate b.

Pre-Algebra

10-5 Solving for a Variable

Solve for the indicated variable.

Try This: Example 1A

A. Solve y – b + 3 = c for y.

y – b + 3 = c

+ b – 3 + b – 3 Add b and subtract 3 from both sides.

y = c + b – 3 Isolate y.

Pre-Algebra

10-5 Solving for a Variable

Solve for the indicated variable.

Try This: Example 1B

B. Solve p – w + 4 = f for w.

p – w + 4 = f– p – 4 – p – 4 Subtract p and 4 from both

sides.

–w = f – p – 4 Isolate w.

Multiply both sides by –1.–1 (–w) = –1 (f – p – 4)

w = –f + p + 4 Isolate w.

Pre-Algebra

10-5 Solving for a Variable

To isolate a variable, you can multiply or divide both sides of an equation by a variable if it can never be equal to 0. You can also take the square root of both sides of an equation that cannot have negative values.

Pre-Algebra

10-5 Solving for a Variable

Solve for the indicated variable. Assume all values are positive.

Additional Example 2A: Solving for a Variable by Division or Square Roots

A. Solve A = s2 for s.

A = s2

√A = s Isolate s.

√A = √s2 Take the square root of both sides.

Pre-Algebra

10-5 Solving for a VariableAdditional Example 2B: Solving for a Variable by Division

or Square Roots

B. Solve V = IR for R.

V = IR

= Divide both sides by I.IRI

VI

IV= R Isolate R.

Solve for the indicated variable. Assume all values are positive.

Pre-Algebra

10-5 Solving for a VariableAdditional Example 2C: Solving for a Variable by Division

or Square Roots

C. Solve the formula for the area of a trapezoid for h. Assume all values are positive.

A = h(b1 + b2)12 Write the formula.

2 • A = 2 • h(b1 + b2)12 Multiply both sides by 2.

2A = h(b1 + b2)

2A(b1 + b2)

= h Isolate h.

2A(b1 + b2)

h(b1 + b2) (b1 + b2)

= Divide both sides by (b1 + b2).

Pre-Algebra

10-5 Solving for a Variable

Solve for the indicated variable. Assume all values are positive.

Try This: Example 2A

A. Solve w = g2 + 4 for g.

w = g2 + 4

√ w – 4 = g Isolate g.

√w – 4 = √g2

w – 4 = g2

– 4 – 4 Subtract 4 from both sides.

Take the square root of both sides.

Pre-Algebra

10-5 Solving for a Variable

B. Solve A = lw for w.

A = lw

= Divide both sides by l.lwl

Al

lA= w Isolate w.

Solve for the indicated variable. Assume all values are positive.

Try This: Example 2B

Pre-Algebra

10-5 Solving for a Variable

C. Solve s = 180(n – 2) for n.

s = 180(n – 2)

Solve for the indicated variable. Assume all values are positive.

Try This: Example 2C

s 180(n – 2)180 180

= Divide both sides by 180.

s180

= (n – 2)

Add 2 to both sides.

s180

+ 2 = n

+ 2 + 2

Pre-Algebra

10-5 Solving for a Variable

To find solutions (x, y), choose values for x substitute to find y.

Remember!

Pre-Algebra

10-5 Solving for a Variable

Additional Example 3: Solving for y and Graphing

Solve for y and graph 3x + 2y = 8.

3x + 2y = 8

–3x –3x

2y = –3x + 8

–3x + 82

2y2

=

y = + 4–3x2

x y

–2 7

0 4

2 1

4 –2

Pre-Algebra

10-5 Solving for a Variable

Additional Example 3 Continued

3x + 2y = 8

Pre-Algebra

10-5 Solving for a Variable

Try This: Example 3

Solve for y and graph 4x + 3y = 12.

4x + 3y = 12

–4x – 4x

3y = –4x + 12

–4x + 12 3

3y3

=

y = + 4–4x3

x y

–3 8

0 4

3 0

6 –4

Pre-Algebra

10-5 Solving for a Variable

Try This: Example 3 Continued

x

y

–4 –2 2 4 6

10

86

2

–2

–4

44x + 3y = 12

–6

Pre-Algebra

10-5 Solving for a Variable

Lesson Quiz: Part 1

Solve for the indicated variable.

1. P = R – C for C.

2. P = 2l+ 2w for l.

3. V = Ah for h.

4. R = for S.

C = R - P

C – Rt = S

13C – S

t

= h3VA

= lP – 2w2

Pre-Algebra

10-5 Solving for a Variable

Lesson Quiz: Part 2

5. Solve for y and graph 2x + 7y = 14.

y = – + 2 2x 7